JEE Main 2023TrigonometryTrigonometric Ratio and IdentitesHardQuestionIf cotα\alphaα = 1 and secβ\betaβ = −53 - {5 \over 3}−35, where π<α<3π2\pi < \alpha < {{3\pi } \over 2}π<α<23π and π2<β<π{\pi \over 2} < \beta < \pi 2π<β<π, then the value of tan(α+β)\tan (\alpha + \beta )tan(α+β) and the quadrant in which α\alphaα + β\betaβ lies, respectively are :OptionsA−17 - {1 \over 7}−71 and IV th quadrantB7 and I st quadrantC−-−7 and IV th quadrantD17 {1 \over 7}71 and I st quadrantCheck AnswerHide SolutionSolution∵cotα=1,α∈(π,3π2)\because \cot \alpha=1, \quad \alpha \in\left(\pi, \frac{3 \pi}{2}\right)∵cotα=1,α∈(π,23π) then tanα=1\tan \alpha=1tanα=1 and secβ=−53,β∈(π2,π)\sec \beta=-\frac{5}{3}, \quad \beta \in\left(\frac{\pi}{2}, \pi\right)secβ=−35,β∈(2π,π) then tanβ=−43\tan \beta=-\frac{4}{3}tanβ=−34 ∴tan(α+β)=tanα+tanβ1−tanα⋅tanβ\therefore \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}∴tan(α+β)=1−tanα⋅tanβtanα+tanβ =1−431+43=−17\begin{aligned} &=\frac{1-\frac{4}{3}}{1+\frac{4}{3}} \\\\ &=-\frac{1}{7} \end{aligned}=1+341−34=−71 α+β∈(3π2,2π) i.e. fourth quadrant \alpha+\beta \in\left(\frac{3 \pi}{2}, 2 \pi\right) \text { i.e. fourth quadrant }α+β∈(23π,2π) i.e. fourth quadrant